QUESTION 1
The given system of equation is
[tex]3x + 2y = 7 - - - (1)[/tex]
and
[tex]x - y + 3 = 0 - - - (2)[/tex]
The question requires that, we make x the subject in equation (2) and put it inside equation (1).
So let us express x in terms of y in equation (2) and call it equation (3) to get,
[tex]x = y - 3 - - - (3)[/tex]
We now substitute equation (3) in to (1) to obtain,
[tex]3(y - 3) + 2y = 7[/tex]
Therefore the correct answer is B.
QUESTION 2
The given equations are
[tex]2x + y = 7 - - - (1)[/tex]
[tex]y - x = 1 - - - (2)[/tex]
First let us make all the four possible substitutions.
The first is to make y the subject in equation (2) and substitute in to equation (1) to get,
[tex]2x + x+ 1 = 7[/tex]
The second one is to make x the subject in equation (2) and put in to equation (1) to get,
[tex]2(y - 1) + y = 7[/tex]
The third one is to make y the subject in equation (1) and put it into equation (2) to get,
[tex]7 - 2x - x = 1[/tex]
The fourth one is to make x the subject in equation (1) and put it in to equation (2) to get,
[tex]y - \frac{(7 - y)}{2} = 1[/tex]
By comparing to given options, C is not part of the four possible results.
Therefore the correct answer is option C.
QUESTION 3
The equations are
[tex]8x = 2y + 5- - - (1)[/tex]
[tex]3x = y + 7- - - (2)[/tex]
We make y the subject in equation (2) to get,
[tex]y = 3x - 7 - - - (3)[/tex]
We substitute equation (3) in to equation (1) to get,
[tex]8x = 2(3x - 7) + 5[/tex]
We expand to obtain,
[tex]8x = 6x - 14+ 5[/tex]
We group like terms to get,
[tex]8x - 6x = - 14+ 5[/tex]
This implies that,
[tex]2x = - 9[/tex]
Therefore,
[tex]x = - \frac{9}{2} [/tex]
We substitute this value into equation (3) to get,
[tex]y = 3( - \frac{9}{2}) - 7 [/tex]
This implies that,
[tex]y = - \frac{27}{2} - 7 [/tex]
This gives us,
[tex]y = \frac{ - 27 - 14}{2} [/tex]
[tex]y = \frac{ - 41}{2} [/tex]
The solution set is
{[tex]( - \frac{9}{2} , - \frac{41}{2} )[/tex]}
The correct answer is B.
